Linearly equivalent divisors and linear transformations

Question

Let $C$ be a projective nonsingular irreducible curve. Let $D$ be a divisor on $C$. Suppose $l(D) = n > 0$ and $L(D) = \langle f_1, \ldots, f_n \rangle$. Consider the map $\varphi_D : C \to \mathbb{P}^{n - 1}$ defined by $\varphi_D(P) = (f_1(P) : \cdots : f_n(P))$. If $D$ and $D'$ are linearly equivalent divisors, does there exist a projective linear transformation $\psi : \mathbb{P}^{n - 1} \to \mathbb{P}^{n - 1}$ such that $\varphi_{D'}(C) = \psi(\varphi_D(C))$?

Answer 1

As written the question doesn't quite make sense, because $\varphi_{D'}$ isn't defined until you specify a basis for $L(D')$. Let me convince you that with the most obvious choice of basis for $L(D')$, the two maps $\varphi_D$ and $\varphi_{D'}$ are actually equal.

Say $D' = D + \operatorname{div} (g)$ for a nonzero rational function $g$.

Then \begin{align*}f \in L(D) &\iff \operatorname{div}(f)+D \geq 0 \\ &\iff \operatorname{div}(f)-\operatorname{div} (g) + D' \geq 0 \\ &\iff \operatorname{div}(f/g) + D' \geq 0 \\ &\iff f/g \in L(D'). \end{align*}

So $(f_1/g,\ldots, f_n/g)$ is a basis for $L(D')$.

The map to projective space given by this basis is

$$\varphi_{D'}(P) = \left( \frac{f_1}{g}(P) : \cdots : \frac{f_n}{g} (P) \right)$$

But multiplying through by $g$, this is nothing other than

$$ \left( f_1(P) : \cdots : f_n(P) \right) = \varphi_D(P).$$